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3 years ago

15

A television station allows 15 minutes of advertising each hour. How many 30 second (1/2 minute) commercials can be run in 1 day

?

1 answer:

AleksAgata [21]3 years ago

3 0

(15 minutes/hour) x (1 spot / 1/2 minute) x (24 hour/day) = <u><em>720</em></u> spots/day

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Which graph represents the solution set to the system of inequalities?<br><br> y≤1/4 x−2, y≥−5/4 x+2

jenyasd209 [6]

The first one. <_ means below and >_ means above. <_ and >_ means solid line < and > mean dotted line. i’m learning this in math right now

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3 years ago

Two miners are trying to push a stationary mine cart filled with gold. Miner A is exerting 25 N while miner B is exerting 20 N.

Oliga [24]

Answer:

b. 45 N

Step-by-step explanation:

The sum of the forces exerted by the two miners = 25 + 20

= 45 N

Since after applying a force of 45 N by the two miners the cart remained stationary, it implies that a resistance force of at least 45 N acts against their push. Thus, the force exerted by the road could be equal to 45 N or greater than 45 N.

Probably, if the sum of the forces applied is 46 N, the cart would move. Which depends on the value of resistance force applied by the road. Thus, the appropriate answer from the given option is B.

Since the force applied on the cart equals the resistance force exerted by the road, the cart would not move.

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2 years ago

50 points to whoever can answer!

Pavlova-9 [17]

Answer:

D

Step-by-step explanation:

I think I'm not sure about that

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3 years ago

Write parametric equations of the line through the points (7,1,-5) and (3,4,-2). please use the first point as your base-point w

Roman55 [17]

Given:

A line through the points (7,1,-5) and (3,4,-2).

To find:

The parametric equations of the line.

Solution:

Direction vector for the points (7,1,-5) and (3,4,-2) is

$\vec {v}=\left$

$\vec {v}=\left$

$\vec {v}=\left$

Now, the perimetric equations for initial point $(x_0,y_0,z_0)$ with direction vector $\vec{v}=\left$ , are

$x=x_0+at$

$y=y_0+bt$

$z=z_0+ct$

The initial point is (7,1,-5) and direction vector is $\vec {v}=\left$ . So the perimetric equations are

$x=(7)+(-4)t$

$x=7-4t$

Similarly,

$y=1+3t$

$z=-5+3t$

Therefore, the required perimetric equations are $x=7-4t, y=1+3t$ and $z=-5+3t$ .

5 0

3 years ago

It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir

Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.95}{2} = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.025 = 0.975$ , so $z = 1.96$

Now, we find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

$\sigma = \sqrt{484} = 22$

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

$M = z*\frac{\sigma}{\sqrt{n}}$

$5 = 1.96*\frac{22}{\sqrt{n}}$

$5\sqrt{n} = 43.12$

$\sqrt{n} = \frac{43.12}{5}$

$\sqrt{n} = 8.624$

$(\sqrt{n})^{2} = (8.624)^{2}$

$n = 74.4$

We need a sample size of at least 75.

6 0

3 years ago

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